Problem: The grades on a language midterm at Almond are normally distributed with $\mu = 84$ and $\sigma = 4.5$. Stephanie earned a n $80$ on the exam. Find the z-score for Stephanie's exam grade. Round to two decimal places.
Solution: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Stephanie's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{80 - {84}}{{4.5}}} $ ${ z \approx -0.89}$ The z-score is $-0.89$. In other words, Stephanie's score was $0.89$ standard deviations below the mean.